[/quote]
,
![\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]](http://latex.artofproblemsolving.com/5/6/b/56b8cb8c13616501dd8308989cea2b338e91598b.png)
,
,
,
be four points in the plane, with 
,
and 
.![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles 
with respect to 
is greater than 0 if 
and since 
?
be an acute scalene triangle with circumcircle 
.
.
and 
be the projections of 
,
,
,
.
.
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
.
for every 
.
,
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide 
,
.
,
,
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
such that 
coprime, 
and 
.
Y by Danielzh, rickyjwalland, jocaleby1
Every morning, Aya does a 
kilometer walk, and then finishes at the coffee shop. One day, she walks at 
kilometers per hour, and the walk takes 
hours, including 
minutes at the coffee shop. Another morning, she walks at 
kilometers per hour, and the walk takes 
hours and 
minutes, including minutes at the coffee shop. This morning, if she walks at 
kilometers per hour, how many minutes will the walk take, including the minutes at the coffee shop?
kilometer walk, and then finishes at the coffee shop. One day, she walks at 
kilometers per hour, and the walk takes 
hours, including 
minutes at the coffee shop. Another morning, she walks at 
kilometers per hour, and the walk takes 
hours and 
minutes, including minutes at the coffee shop. This morning, if she walks at 
kilometers per hour, how many minutes will the walk take, including the minutes at the coffee shop?
.
.
,
.
.
.
hours, which corresponds to 
.![\[60 \cdot \frac{9}{.5 + 2.5} + 24 = \boxed{204}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/a/6/7/a6751bb1aaa59f58c68a9219ab6992fed5219d5b.png)
km/h, Aya is able to walk 
hours. Therefore, we have the system
Subtracting the second from the first gives 
.
,
.
hours on the run itself, so
Similarly, we get that 
Subtracting these gives us that
or
hence
Substituting this back into one of the equations, we find that 
and 
so we get 
minutes as our answer.
=
.
=
.
=
,
=
.
=
,
=
= 
hr-t. So, 
and 
=
.
.
=
,
.
,
Subtract the two then we get:
By taking the first equation, we can solve for 
.
we get 
hours which is 
and 
,![\[96=540\left(\frac{1}{s}-\frac{1}{s+2}\right)=\frac{540\cdot2}{s(s+2)}\implies s(s+2)=\frac{45}{4}\implies s=\frac{5}{2}.\]](http://latex.artofproblemsolving.com/2/b/2/2b2b50ab8fb976b597f72d7d052fcac516275219.png)
Thus 
and the answer is 
.
since 
.
.
,
(in minutes). This leads us to get an answer of 
.
